Answer:
The enthalpy of formation of CaF₂ is -1224.4 kJ.
Explanation:
The enthalpy of formation of CaF₂ can be calculated as follows:
[tex] \Delta H_{f} = \frac{q}{n_{CaF_{2}}} [/tex]
Where:
q: is the heat liberated in the reaction = -251 kJ
The number of moles of CaF₂ is:
[tex] n_{CaF_{2}} = \frac{m}{M} [/tex]
Where:
m: is the mass of CaF₂ = 16 g
M: is the molar mass of CaF₂ = 78.07 g/mol
[tex] n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles [/tex]
Now, the enthalpy of formation of CaF₂ is:
[tex]\Delta H_{f} = \frac{q}{n_{CaF_{2}}} = \frac{-251 \cdot 10^{3} J}{0.205 moles} = -1224.4 kJ/mol[/tex]
Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.
I hope it helps you!
